Question

How do you find the critical numbers of `y= x^2*(1 + 3 ln x)`?

Answer 1

`x=e^(-5/6)`

Explanation:

Differentiate, set equal to zero, and solve for `x,` in addition to being aware of specific values for which the derivative doesn't exist.

`y'=x^2*d/dx(1+3lnx)+(1+3lnx)*d/dxx^2`

`y'=x^2(3/x)+2x(1+3lnx)`

`y'=3x+2x+6xlnx`

`y'=5x+6xlnx`

The derivative is logarithmic like the original function. They both share the domain `(0, oo),` so there aren't any particular values for which the derivative doesn't exist while the function does.

`5x+6xlnx=0`

`x(5+6lnx)=0`

`x=0` looks like it would be a solution, but since it's not in the domain of the logarithm, it is not valid.

`5+6lnx=0`

`6lnx=-5`

`lnx=-5/6`

`e^lnx=e^(-5/6)`

From the fact that `e^lnx=x:`

`x=e^(-5/6)`

Is the only critical number.